/*
	Faster and clearer!
	
	d[n] = number of different ways to produce n cents.
	for j = 1 -> 5
		for i = 1 -> n
			d[i] += d[i-c[j]] 
*/

#include <iostream>

using namespace std;

int c[5] = {1, 5, 10, 25, 50};
long long d[30001];
int n;

int main()
{
	d[0] = 1;
	for (int j = 0; j < 5; j++)
		for (int i = 1; i <= 30000; i++)
			if (i - c[j] >= 0) d[i] += d[i-c[j]];
	
	while (cin >> n)
	{
		if ((n == 0) || (d[n] == 1))
			cout << "There is only 1 way to produce " << n << " cents change." << endl;
		else
			cout << "There are " << d[n] << " ways to produce " << n << " cents change." << endl;
	}

	return 0;
}
